Question

cho sinα = \(\dfrac{1}{4}\)

tính B = \(\dfrac{3\cot\alpha-tan\alpha}{2tan\alpha+cot\alpha}\)

Answer 1

sin a=1/4

=>sin^2a=1/16

=>cos^2a=15/16

\(B=\dfrac{3\cdot\dfrac{cosa}{sina}-\dfrac{sina}{cosa}}{2\cdot\dfrac{sina}{cosa}+\dfrac{cosa}{sina}}\)

\(=\dfrac{3\cdot cosa^2a-sin^2a}{sina\cdot cosa}:\dfrac{2\cdot sin^2a+cos^2a}{sina\cdot cosa}\)

\(=\dfrac{3\cdot cos^2a-sin^2a}{2\cdot sin^2a+cos^2a}\)

\(=\dfrac{3\cdot\dfrac{15}{16}-\dfrac{1}{16}}{2\cdot\dfrac{1}{16}+\dfrac{15}{16}}=\dfrac{44}{17}\)